# Derivation of Mean Thermal Wavelength

1. Aug 14, 2006

### GreenPenInc

Hi guys,

I'm trying to derive the mean thermal wavelength from the Maxwell distribution:

$$M(v) = 4\pi\left(\frac{m}{2\pi k_BT}\right)^{3/2}v^2e^{-\frac{mv^2}{2k_BT}} = 4\pi\left(\frac{a}{\pi}\right)^{3/2}v^2e^{-av^2}$$

With $$a = \frac{m}{2k_BT}$$ introduced for convenience. Since $$\lambda = \frac{h}{mv}$$, I figured an expression for the thermal wavelength would be

$$\lambda_T = \frac{h}{m}\int_0^\infty\frac{M(v)}{v}dv$$

Problem is, when I do this, I end up with

$$\lambda_T = 2\sqrt{\frac{h^2}{2\pi m k_BT}}$$

This is exactly twice the accepted value. Where did I go wrong?

2. Aug 14, 2006

### Physics Monkey

Hi GreenPenInc,

Don't worry too much about the factor of two in your answer. The calculation you have made is perfectly reasonable and physical; as it must, it reproduces the standard definition up to a factor of order one. The reason why you don't get the standard definition is that the thermal de Broglie wavelength is defined in a slightly different fashion. In fact, the factors of pi and such are chosen so that the equation of state for the ideal Bose gas (and the ideal Fermi gas) takes a particularly simple form. When you put all the factors in you find that the pressure of an ideal Bose gas is given by
$$\frac{P}{k T} = \frac{g}{\lambda_T^3} g_{5/2}(z),$$
while for a Fermi gas the result is
$$\frac{P}{k T} = \frac{f}{\lambda_T^3} f_{5/2}(z).$$
The meaning of the symbols on the left you certainly know; the factor $$g$$ is called the degeneracy, and it usually measures something like the number of spin components or polarization states, etc. The variable $$z = e^{\beta \mu}$$ is called the fugacity, and it depends in a simple way on the chemical potential $$\mu$$. The functions $$g_n$$ and $$f_n$$ are called Bose and Fermi functions, respectively. They are certain special functions that appear again and again in the theory of ideal quantum gases. For example, the definition of $$g_n(z)$$ is $$g_n(z) = \frac{1}{\Gamma(n)} \int^\infty_0 \frac{x^{n-1}}{z^{-1} e^x - 1} dx$$. If you haven't already, you will meet these functions in a course on quantum statistical mechanics.

So, to bring this long winded answer to a close, the thermal de Broglie wavelength is defined so that the equation of state looks nice and neat. Of course, the physics doesn't care about factors of 2 and pi, so your choice would work just as well. Hope this helps.

Last edited: Aug 14, 2006
3. Aug 14, 2006

### GreenPenInc

This was essentially the answer which a prof I know gave when several of us asked him today, minus the reason for it taking the specific form it takes (he didn't know off the top of his head). That is very enlightening. I remember those f and g functions, although I learned them in a slightly generalized form $$h_\nu(z; a)$$, where the parameter a could be adjusted to deal with bosons, fermions, or the fictitious "boltzons" (classical limit).

I'm studying for the quals right now, which is how this came up in the first place, and this is all starting to come back to me.

Now that I know the rationale, it annoys me to see it referred to as the "mean thermal wavelength" in so many textbooks; I think it would be better to call it something like the "characteristic thermal wavelength".

Thanks again!

4. Aug 14, 2006

### Physics Monkey

Glad to hear I could help, and good luck with your quals!

P.S. Welcome to the forums.

5. Sep 16, 2006

### GreenPenInc

Almost forgot about this thread. Turns out I passed my quals with flying colours, although almost 2/3 of the people failed the second day (modern physics). All that studying paid off. :)